Question

use an appropriate half - angle formula to find the exact value of the expression. $sin(\frac{11pi}{12})$

Explanation:

Step1: Rewrite the angle

We know that $\frac{11\pi}{12}=\frac{\frac{11\pi}{6}}{2}$.

Step2: Recall the half - angle formula for sine

The half - angle formula for sine is $\sin\frac{\alpha}{2}=\pm\sqrt{\frac{1 - \cos\alpha}{2}}$. Here $\alpha=\frac{11\pi}{6}$.

Step3: Find the value of $\cos\alpha$

$\cos\frac{11\pi}{6}=\cos(2\pi-\frac{\pi}{6})=\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}$.

Step4: Determine the sign of $\sin\frac{11\pi}{12}$

Since $\frac{11\pi}{12}$ is in the second quadrant, $\sin\frac{11\pi}{12}>0$.

Step5: Substitute into the half - angle formula

$\sin\frac{11\pi}{12}=\sqrt{\frac{1-\cos\frac{11\pi}{6}}{2}}=\sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}}=\sqrt{\frac{2-\sqrt{3}}{4}}=\frac{\sqrt{2 - \sqrt{3}}}{2}$.

Answer:

$\frac{\sqrt{2-\sqrt{3}}}{2}$